Here you will find daily news and tutorials about R, contributed by over 750 bloggers.
There are many ways to follow us - By e-mail:On Facebook: If you are an R blogger yourself you are invited to add your own R content feed to this site (Non-English R bloggers should add themselves- here)

Facing a simple, yet frustrating formula like this

and the task to solve it for x left me googling around for hours until I found salvation in Wolfram Alpha, Wikipedia, and a nice blog post with R-syntax to solve a similar equation.

Using the results from Wolfram Alpha I was able to find the solution with the ‘gsl’ library

# install.packages("gsl")
library(gsl)
# create some example data
dat <- data.frame(a = 0.109861, x = 10)
# a is set so that b is roughly 30.
# Lazy as I am I used Excel and its solver ability to find numbers
# to check if b is close to 30. Using the initial formula
dat$b <- dat$x * exp(dat$a * dat$x)
dat
# solve for x2 and see if x and x2 are similar and close to 10
dat$x2 <- lambert_W0(dat$a * dat$b)/dat$a
dat
# a x b2 x2
#1 0.109861 10 29.99993 10.00001
# Hurray!

Sometimes life can be so easy (after a long time searching for the right results….).

Related

To leave a comment for the author, please follow the link and comment on their blog: Data Shenanigans » R.